Problem

The current in a circuit falls from $5.0 \, \mathrm{A}$ to $0 \, \mathrm{A}$ in $0.1 \, \mathrm{s}$. If an average e.m.f. of $200 \, \mathrm{V}$ is induced, estimate the self-inductance of the circuit.

Data

• Average e.m.f.: $\varepsilon = 200 \, \mathrm{V}$
• Initial current: $I_i = 5.0 \, \mathrm{A}$
• Final current: $I_f = 0 \, \mathrm{A}$
• Time interval: $\Delta t = 0.1 \, \mathrm{s}$

To find:

• Self-inductance $L$.

Prerequisite Concepts

1. The relation for self-inductance:

$$L = \frac{\varepsilon \Delta t}{\Delta I},$$

where:

• $\varepsilon$ is the average induced e.m.f.,
• $\Delta I = I_f - I_i$ is the change in current,
• $\Delta t$ is the time taken for the change.

Solution

1. Calculate the change in current:

$$\Delta I = I_f - I_i = 0 - 5 = -5 \, \mathrm{A}.$$

(The negative sign indicates a decrease in current.)

2. Substitute into the formula for self-inductance:

$$L = \frac{\varepsilon \Delta t}{\Delta I}.$$

3. Substitute the given values:

$$L = \frac{200 \times 0.1}{5}.$$

4. Simplify:

$$L = \frac{20}{5} = 4 \, \mathrm{H}.$$

Answer

The self-inductance of the circuit is:

$$L = 4 \, \mathrm{H}.$$

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